Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N;
double sy, xa, xb;
pair<double, double> seg[200010];
double csum[200010];
double getLength(double x, double y) {
  double pa = x - y * (x - xa) / (y + sy), pb = x + y * (xb - x) / (y + sy);
  int lo = upper_bound(seg + 1, seg + N + 1, make_pair(pa, 0.0)) - seg - 1;
  int hi = upper_bound(seg + 1, seg + N + 1, make_pair(pb, 0.0)) - seg - 1;
  double total = csum[hi] - csum[lo - 1];
  if (lo != 0) total -= min(seg[lo].second - seg[lo].first, pa - seg[lo].first);
  if (hi != 0 && pb >= seg[hi].first && pb <= seg[hi].second)
    total -= seg[hi].second - pb;
  return total;
}
int main() {
  scanf("%lf %lf %lf %d", &sy, &xa, &xb, &N);
  sy = -sy;
  for (int i = 1; i <= N; i++) scanf("%lf %lf", &seg[i].first, &seg[i].second);
  sort(seg + 1, seg + N + 1);
  for (int i = 1; i <= N; i++) csum[i] = seg[i].second - seg[i].first;
  for (int i = 1; i <= N; i++) csum[i] += csum[i - 1];
  int Q;
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    double x, y;
    scanf("%lf %lf", &x, &y);
    double ans = getLength(x, y);
    ans = ans * (1 + sy / y);
    printf("%0.9lf\n", ans);
  }
  return 0;
}